Trigonometry puzzling problem

Question

Contructed the figure here reported and known that $AB=2r$, $AC=r\sqrt{2}$ and fixed $x:=P\hat{A}C$, it is asked to find $$ y = CK + PH \sqrt{2} + PK $$

as a function of $x$.

I started with

$$PH = AP\sin x$$

but I can't see how to see the other lengths missing.

Can anyone give me some suggestions? I don't know how to go on....

Thanks.

Answer 1

The lines $PK$ and $KC$ both touch the circle. So if you fill the second half of the circle and add two more orthogonal lines, the circle will be boxed in a square. The points $P$ and $C$ will be on the centres of the square's edges. So $PK = KC = r$.

Answer 2

First of all, $AC=\sqrt2 r$ entails $\angle CAB=45°$, and consequently $\angle AOP=90°-2x$ and $\angle COP=2x$, where $O$ is the center of the circle (midpoint of $AB$). That is enough to compute $y$.